# In triangle ABC, prove the following:

Question:

In triangle ABC, prove the following:

$\frac{c}{a-b}=\frac{\tan \left(\frac{A}{2}\right)+\tan \left(\frac{B}{2}\right)}{\tan \left(\frac{A}{2}\right)-\tan \left(\frac{B}{2}\right)}$

Solution:

Let $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$   ...(1)

We need to prove:

$\frac{c}{a-b}=\frac{\tan \left(\frac{A}{2}\right)+\tan \left(\frac{B}{2}\right)}{\tan \left(\frac{A}{2}\right)-\tan \left(\frac{B}{2}\right)}$

Consider

$\mathrm{LHS}=\frac{c}{a-b}$

$=\frac{k \sin C}{k(\sin A-\sin B)} \quad$ (using (1))

$=\frac{2 \sin \frac{C}{2} \cos \frac{C}{2}}{2 \sin \left(\frac{A-B}{2}\right) \cos \left(\frac{A+B}{2}\right)}$

$=\frac{\sin \left(\frac{\pi-(A+B)}{2}\right) \cos \frac{C}{2}}{\sin \left(\frac{A-B}{2}\right) \cos \left(\frac{A+B}{2}\right)} \quad(\because \mathrm{A}+\mathrm{B}+\mathrm{C}=\pi)$

$=\frac{\cos \frac{C}{2} \cos \left(\frac{A+B}{2}\right)}{\sin \left(\frac{A-B}{2}\right) \cos \left(\frac{A+B}{2}\right)}$

$=\frac{\cos \frac{C}{2}}{\sin \left(\frac{A-B}{2}\right)} \quad \ldots(2)$

RHS $=\frac{\tan \frac{A}{2}+\tan \frac{B}{2}}{\tan \frac{A}{2}-\tan \frac{B}{2}}$

$=\frac{\frac{\sin \frac{A}{2}}{\cos \frac{A}{2}}+\frac{\sin \frac{B}{2}}{\cos \frac{B}{2}}}{\frac{\sin \frac{A}{2}}{\cos \frac{A}{2}}-\frac{\sin \frac{B}{2}}{\cos \frac{B}{2}}}$

$=\frac{\sin \frac{A}{2} \cos \frac{B}{2}+\sin \frac{B}{2} \cos \frac{A}{2}}{\sin \frac{A}{2} \cos \frac{B}{2}-\sin \frac{B}{2} \cos \frac{A}{2}}$

$=\frac{\sin \left(\frac{A+B}{2}\right)}{\sin \left(\frac{A-B}{2}\right)}$

$=\frac{\sin \left(\frac{\pi-\mathrm{C}}{2}\right)}{\sin \left(\frac{A-B}{2}\right)}$

$=\frac{\cos \frac{C}{2}}{\sin \left(\frac{A-B}{2}\right)}$

$=\mathrm{LHS} \quad($ from $(2))$

Hence proved.